Question

A tower subtends angles $\alpha ,2\alpha$ and $3\alpha$ respectively at points, $A,B$ and $C$ (all points lying on the same side on a horizontal line through the foot of the tower), then the value of $\frac{AB}{BC}$ is equal to

• A. $1+2cos2\alpha$
• B. $1-2cos2\alpha$
• C. $1+3cos2\alpha$
• D. $1-3cos2\alpha$

Answer 1

Answer: (A)

Let, the height of the tower is $PQ=h$
Where, $Q$ is the foot of the tower and $P$ is the top of the tower, then
$QA=hcot\alpha$
$QB=hcot2\alpha$
$QC=hcot3\alpha$

$\frac{AB}{BC}=\frac{QA-QB}{QB-QC}=\frac{cot\alpha -cot2\alpha }{cot2\alpha -cot3\alpha }$
$=\frac{\frac{cos\alpha }{sin\alpha }-\frac{cos2\alpha }{sin2\alpha }}{\frac{cos2\alpha }{sin2\alpha }-\frac{cos3\alpha }{sin3\alpha }}=\frac{sin2\alpha cos\alpha -cos2\alpha sin\alpha }{sin3\alpha cos2\alpha -cos3\alpha sin2\alpha }\times \frac{sin3\alpha }{sin\alpha }$
$=\frac{sin3\alpha }{sin\alpha }=3-4si{n}^2\alpha =1+2cos2\alpha$